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QUANT TECHNIQUES STRAIGHT FROM SERIAL CAT TOPPER BYJU INDEX

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1) CONSTANT PRODUCT RULE(1/x) & 1/(x+1)

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1) CONSTANT PRODUCT RULE 2) POWER CYCLE 3) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS 4) LAST 2 DIGITS TECHNIQUE 5) SIMLIAR TO DIFFERENT GROUPING ( P&C) 6) APPLICATION OF FACTORIALS 7) GRAPHICAL DIVISION FOR GEOMETRY 8) ASSUMPTION METHOD

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This rule can be applied when we have two parameters whose product is constant, or in other words, when they are inversely proportional to each other. eg)Time x Speed = Distance, Price x consumption = Expenditure, Length x breadth= Area A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other Parameter if the parameters are inversely proportional. Lets see the application with the following examples

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1) A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office? Solution: Assume original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or ½. 1/2 increase in speed will result in 1/3 decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5 hoursà Answer is Distance= 4x1.5=6 km

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2) A 20% increase in price of sugar. Find the % decrease in consumption a family should adopt so that the expenditure remains constant Solution: Here price x consumption= expenditure (constant) Using the technique, 20% (1/5) increase results in 16.66%(1/6) decrease in consumption. Answer=16.66%

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2) POWER CYCLE The last digit of a number of the form ab falls in a particular sequence or order depending on the unit digit of the number (a) and the power the number is raised to (b). The power cycle of a number thus depends on its’ unit digit. Consider the power cycle of 2 2 1=2, 2 5=32 2 2 =4 2 6=64 3 2 =8 2 7=128 4 2 =16 2 8=256 As it can be observed, the unit digit gets repeated after every 4 th power of 2. Hence, we can say that 2 has a power cycle of 2,4,8,6 with frequency 4. This means that, a number of the form 2 4k+1 will have the last digit as 2 2 4k+2 will have the last digit as 4 2 4k+3 will have the last digit as 8 2 4k+4 will have the last digit as 6 (where k=0, 1, 2, 3…) This is applicable not only for 2, but for all numbers ending in 2. Therefore to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0 to 9, which are given below

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Power cycle 0 1 2,4,8,6 3,9,7,1 4,6 5 6 7,9,3,1 8,4,2,6 9,1

Frequency 1 1 4 4 2 1 1 4 4 2

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Unit digit 0 1 2 3 4 5 6 7 8 9

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For example 3) Find the remainder when 375 is divided by 5. 1) Express power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3. 2) Take the power cycle of 3 which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7 Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2. Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in the fastest way possible. For example,

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4) Find the unit digit of 73^4n Put n=1, the problem reduces to 7 3^4, which is 781. Since 81=4k+1, take the first digit in the power cycle of 7, which is 7.

What is the first non zero integer from the right in 83301957 + 8370 1982? a) 3 b) 1 c) 9 d) none of these 8370 1982 will end with more number of zeroes so we need to consider only the first part. Rightmost non-zero integer of the expression will be = unit digit of 8331957 = unit digit of 31957 Since 1957=4k+1, take the first digit in the power cycle of 3, which is 3.

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If N = (13)1! + 2! + 3! + ..+ 13! + (28)1! + 2! + 3!..+ 28! + (32)1! + 2! + 3! + ...+ 32!+ (67) 1! + 2! + 3! + then the unit digit of N is (a) 4 (b) 8 (c) 2 (d) none of these Based on Power Cycle After 4! Every number is of the form 4k+4 , here in all four terms power becomes 4k+1. So taking the first digit from the power cycles of 3,8,2, and 7 we will get the unit digit as i.e 3+8+2+7 = 4. Ans = 0 6)

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3) Useful technique to find the last 2 digits of any expression of the form ab Depending on the last digit of the number in question, we can find the last two digits of that number. We can classify the technique to be applied into 4 categories. TYPE METHOD EXAMPLES 67 1) Numbers ending in 1 The last digit is always 1. 1) 21 =__41(2 * 7=__4) The 2nd last digit = product of tens digit of base * unit digit of the power. In 2167; 2 is the tens digit of base and 7 is the unit digit of power

2) 4187 =__ 81( 4 *7= __8)

2) Number ending with 5

Last two digits: always 25

e.g.) 1555 34 = __ 25

3) Numbers ending in 3, 7, 9

Change the power so that the base ends with 1 and then use the same technique as for those numbers ending with 1.

e.g.) 17288 =(174)72 (taking power 4 as 74 will end in 1. (17 2* 172) 72

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4) 31124 =__ 21( 3 * 4=__2)

=( _89*_89)72 (as last 2 digits of 172=89) = ( _21)72 (as last 2 digits of 89*89=21) Answer =__41( as 2*2=4)

Use the pattern of the number 1024 =2 10 i.e. *2 10 raised to even power ends with 76 and * 210raised to odd power ends with 24.

e.g.) 2 788 = (210) 78 * 28 =

__76 * __56 = __56.

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4) For even numbers (2,4,6,8)

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eg) 34, 7 4 &9 2 all will end in1.

3) 1261 167 =__21 ( 6 * 7= __2)

It is also important to note that,

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1. 76 multiplied by the last 2 digits of any power of 2 will end in the same last 2 digits E.g. 76*04 = 04, 76*08 = 08, 76*16 = 16, 76*32 = 32 2. The last two digits of , x2, (50-x)2 , (50+x)2 , (100-x)2 will always be the same. For example last 2 digits of 122,382,622,882,112 2…. will all be the same. Also, last two digits of 11 2=392=612=89 2 =111 2=139 2=161 2=1892 and so on 3.To find the squares of numbers from 30-70 we can use the following method 7) To find 41 2 Step1 : Difference from 25 will be first 2 digits = 16 Step 2 : Square of the difference from 50 will be last 2 digits = 81 Answer = 1681. 8): To find 432 Step1 : Difference from 25 will be first 2 digits = 18 Step 2 :Square of the difference from 50 will be last 2 digits = 49 Answer = 1849

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4. Combining all these techniques we can find the last 2 digits for any number because every even number can be written as 2* an odd number

4) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS 9) In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B, 80 use brand C, 90 use brand D & 60 use brand E of the same product. What is the minimum possible number of men using all the 5 brands, if all the 100 men use at least one of these brands?

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Sum of the difference from 100 = (100-100) + (100-75)+(100-80)+(100-90)+(100-60) = 95 Again take the difference from 100 = 5 (answer) This method will be explained in detail in the next booklet.

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5) SIMILAR TO DIFFERENT GROUPING IN PERMUTATION & COMBINATION

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All questions in Permutation and Combination fall into 4 categories, and if you master these 4 categories, you can understand all concepts in P&C easily. 1) Similar to Different 2) Different to Similar 3) Similar to Similar 4) Different to Different In this booklet, we will look at the first category; i.e. Similar to Different, which entails the number of ways of dividing ‘n’ identical (similar) things into ‘r’ distinct (different) groups

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a) NO LIMIT QUESTIONS

Let me explain this with an example. Suppose I have 10 identical chocolates to be divided among 3 people. The 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates.

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So let us represent chocolates by blue balls. The straight red lines are used to divide them into parts. So you can see that for dividing into 3 parts, you need only two lines.

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Suppose you want to give 1 st person 1 chocolate, 2nd 3 chocolates and 3rd 6 chocolates. Then you can show it as:

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Suppose you want to give one person 1 chocolate, another person 6 chocolates and another one 3, then it can be represented as:

Now if first person gets 0, second gets 1 and third gets 9 chocolates then it can be represented as:

Now suppose you want to give first person 0, second also 0 and third all of 10 then you can show it like:

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èSo, for dividing 10 identical chocolates among 3 persons you can assume to have 12 (10 balls +2 sticks) things among which ten are identical and rest 2 are same and of one kind. So number of ways in which you can distribute ten chocolates among 3 people is the same in which you can arrange 12 things among which 10 are identical and of one kind while 2 are identical and of one kind which can be done in 12!/(10!2!)

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The above situation is same as finding the number of positive integral solutions of a + b + c = 10. a, b, c is the number of chocolates given to different persons. Or else I can also say how many integral points lie on the line a + b + c = 10 in the 1st quadrant. In both the cases the answer is 12C2.

b) LOWER LIMIT QUESTIONS

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èNow suppose we have a restriction that the groups cannot be empty i.e. in the above example all 3 persons should get at least 1.

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You have to divide ten chocolates among 3 persons so that each gets at least one. So in the start only give them one each. This you will do in just 1 way as all the chocolates are identical. Now, you are left with 7 chocolates and you have to divide them among 3 people in such that way that each gets 0 or more. You can do this easily as explained above using balls and sticks. Number of ways = 9! /(7!2!).

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The above situation is same as finding the number of positive natural number solutions of a + b + c = 10. a, b, c is the number of chocolates given to different persons. Or else I can also say how many integral natural points lie on the line a + b + c = 10 in the 1 st quadrant. In both the cases the answer is 9C2.

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è Now suppose I change the question and say that now you have to divide 10 chocolates among 3 persons in such a way that one gets at least 1, second at least 2 and third at least 3. Its as simple as the last one. First full fill the required condition. Give 1 st person 1, second 2 and third 3 and then divide the left 4 (10–1–2-3) chocolates among those 3 in such a way that each gets at least 1. This is same as arranging 4 balls and 2 sticks which can be done in 6C2 ways.

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The above situation is same as finding the number of positive integral solutions of a + b + c = 10 such that a > = 1, b > = 2, c > = 3. a, b, c is the number of chocolates given to different persons. In this case the answer is 9C2. 10)

Rajesh went to the market to buy 18 fruits in all. If there were mangoes, bananas, apples and oranges for sale then in how many ways could Rajesh buy at least one fruit of each kind? a) 17C3 b) 18C4 c) 21C3 d) 21C4

This is a Grouping type 1 Similar to Distinct question, with a lower limit condition. M+B+A+O=18 Remove one from each group, therefore 4 is subtracted from both sides. The problem changes to M+B+A+O=14 Using our shortcut, The answer is the arrangement of 14 balls and 3 sticks i.e. 17C3

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11)

There are four balls to be put in five boxes where each box can accommodate any number of balls. In how many ways can one do this if: a) Balls are similar and boxes are different 1) 275 2) 70 c) 120 d) 19

When the balls are similar and the boxes are different, it’s a grouping type 1 question A+B+C+D+E=4, where A,B,C,D,E are the different boxes. The number of ways of selection and distribution=8C4=70 The number of non negative integral solutions of x1+x2+x3 ≤ 10 a) 84 b) 286 c) 220 d) none of these

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By non-negative integral solutions, the conditions imply that we can have 0 and natural number values for x1, x2, x3, and x4 To remove the sign ≤ add another dummy variable x4. The problem changes to x1+x2+x3+x4=10 This is an example of grouping type1 (Similar to Distinct) It is the arrangement of 10 balls and 3 sticks. Using the shortcut of balls and sticks, Therefore the answer is 13C3=286

6) APPLICATION OF FACTORIALS

Definition of Factorialà N! = 1*2*3*…*(n-1)*n Eg 1) 5!= 1*2*3*4*5=120 eg 2) 3!=1*2*3=6

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A thorough understanding of Factorials is important because they play a pivotal role not only in understanding concepts in Numbers but also other important topics like Permutation and Combination

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Let us now look at the application of Factorials

Highest power in a factorial or in a product

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Questions based on highest power in a factorial are seen year after year in CAT. Questions based on this can be categorized based on the nature of the number (prime or composite) whose highest power we are finding in the factorial, i.e

Highest power of a prime number in a factorial:

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To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

13) The highest power of 5 in 100!

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100/5=20; 20/5=4; Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24 ALTERNATIVE METHOD 100/5 + 100/52 =20+4=24 (We take upto 52 as it is the highest power of 5 which is less than 100)

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Highest power of a composite number in factorial

Factorize the number into primes. Find the highest power of all the prime numbers in that factorial using the previous method. Take the least power.

14) To find the highest power of 10 in 100! Solution: Factorize 10=5*2. 1. Highest power of 5 in 100! =24 2. Highest power of 2 in 100! =97 Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. So take the lesser i.e. 24 is the answer.

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15) Highest power of 12 in 100! Solution: 12=22 *3. Find the highest power of 22 and 3 in 100! First find out the highest power of 2. Listing out the quotients:100/2 = 50; 50/2 = 25; 25/2 = 12; 12/2 = 6; 6/2 = 3; 3/2 = 1 Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97. So highest power of 2 2 = 48 (out of 97 2’s only 48 can make 22) Now for the highest power of 3. 100/3 = 33; 33/3 = 11; 11/3 = 3; 3/3 = 1; Highest power of 3 = 48, Highest power of 12 = 48

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Number of zeros in the end of a factorial or a product

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Finding the number of zeroes forms the base concept for a number of application questions. In base 10, number of zeros in the end depends on the number of 10s; i.e. effectively, on the number of 5s In base N, number of zeroes in the end highest power of N in that product

16) Find the number of zeroes in 13! In base 10

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Solution: We need to effectively find the highest power of 10 in 13! = Highest power of 5 in 13! As this power will be lesser. 13/5=2

17) Find the number of zeroes in the end of 15! in base 12.

Number of factors of any factorial

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Solution: Highest power of 12 in 15! =highest power of 22 *3 in 15! =Highest power of 3 in 15!= 5

There is a technique, which can be used to find the number of factors in a factorial

18) Find the factors of 12!

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STEP 1: Prime factorize 12! i.e. find out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! = 2 10*3 5*52*7*11 STEP2: Use the formula N=am*bn (a, b are the prime factors). Then number of factors= (m+1)(n+1) The number of factors= (10+1)(5+1)(2+1)(1+1)(1+1) =792. Answer=792

Right most non zero integer in a factorial

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19) Find the right most non zero integer in 25! ? OR Find the remainder when 25! is divided by 107?( Both the questions are conceptually the same) There are 6 zeroes at the end of 25!. Effectively we need to find the rightmost non-zero integer in 25!. Number of 2s in 25! = 25/2+12/2+6/2+3/2 = 12+6+3+1=22, Number of 3s in 25! = 25/3 + 8/3 = 8+2=10 Number of 5s in 25! = 25/5+5/1 =6, Number of 7s in 25! =25/7=3 Number of 11s in 25! =2, Number of 13s,17s,19s,23s in 25! =1 Remove 26x5 6. (form the 6 zeroes), Right most non zero digit will be last digit of 2163107311 2.131.171.191.231 = 6.9.3.1.3.7.9.3 = __4. Answer is 4.

A shortcut to solve the above question.

If you know the unit digit in the product 1!*2!*3!*……10! = 8 and Unit digit in the product 11!*12*13*……………..20! = 8.then the answer is in one line as follows. 8*8*1*2*3 = Answer 4.This short cut method takes just 30 seconds.

APPLICATION QUESTION BASED ON FACTORIAL 20) How many natural numbers are there such that their factorials are ending with 5 zeroes? 10! is 1*2*3*4*(5)*6*7*8*9*(2*5). From this we can see that highest power of 5 till 10! is 2. Continuing like this, 10!-14!, highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3). Therefore, from 15! To 19! - The highest power of 5 will be 3.

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20!-24! – Highest Power = 4, In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get highest power of 5 as 6. The answer to the question is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

7) USE OF GRAPHICAL DIVISION IN GEOMETRY

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Lets look at a technique which will help you solve a geometry question in no time 21) ABCD is a square and E and F are the midpoints of AB and BC respectively. Find the ratio of Area( ABCD): Area(DEF)?

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8) ASSUMPTION method

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FIGURE A FIGURE B Lets divide the figure using dotted lines as shown in Figure B. Area of ABCD=100%. Area AEDG=50%. Then Area in shaded region 1(AED)= 25%. Similarly, DCFH=50%. Area in shaded region 2(DCF)=25%. Now Area of EOFB= 25%. .Area of shaded region 3(BEF)=12.5%. Total area outside triangle= 62.5%. Area inside triangle= 10062.5=37.5%. required ratio = 100/37.5 = 8:3

There are at least 10 questions out of 25 in CAT 2007 where you can apply this technique. This involves assuming simple values for the variables in the questions, and substituting in answer options based on those values. Assumption helps to tremendously speedup the process of evaluating the answer as shown below.

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22) k & 2k 2 are the two roots of the equation x 2 – px + q. Find q + 4q 2 + 6pq = a) q2 b) p 3 c) 0 d) 2p3 Solution: Assume an equation with roots 1&2 (k=1) =>p (sum of roots)= 3 and q(product of roots)=2. Substitute in q + 4q 2 + 6pq = 54. Look in the answer options for 54 on substituting values of p=3 and q=2. we get 2p3 = 54.=> Ans = 2p3 .

23) Let ‘x’ be the arithmetic mean and y,z be the two geometric means between any two positive numbers. The value of (y3 + z3) / (xyz) is a) 2 b) 3 c) 1/2 d) 3/2 Assume a GP 1 2 4 8 implies 2 GMs between 1 and 8, i.e. y=2 and z=4. Arithmetic Mean, x =(8+1)/2 = 4.5 .Substitute in (y3 + z3) / (xyz) Answer = (2 3+43)/(2x4x4.5) = 2. NOTE:- Assume a GP 1 1 1 1 then x=1, y=1 z=1. Answer on substitution=2, which will make the calculation even faster, half of the problems in Algebra can be solved using assumption. This is not direct substitution. In the next eg: see how you can use the same technique in an equation question.

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SOME ACTUAL QUESTIONS FROM CAT-2007 SOLVED USING THIS TECHNIQUE

c) q(n/2)(p+q)n/2

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25) Let a 1=p and b1=q, p and q are positive quantities. Define a n=pbn-1, b n=qb n-1 for even n>1 a n=pan-1, b n=qa n-1 for odd n>1. which of the following best describes a n+bn for even n. a) qp (n/2)(p+q) b) q(n/2)(p+q)n (n/2)-1 n/2 d)q(pq) (p+q) e)q(pq)(n/2)-1(p+q)

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24) Consider the set S={2,3,4……2n+1), where n is a positive integer larger than 2007. Define X as the average of odd integers in S and Y as the average of the even integers in S. What is the value of X-Y? a) 1 b) n/2 c)(n+1)/2n d) 2008 e) 0 The question is independent of n, which is shown below. Take n=2. Then S= {2,3,4,5). X= 4 and Y=3. X-Y =1, Take n=3. then S={2,3,4,5,6,7}. X=5 and Y=4. X-Y=1 Hence you can directly mark the answer option (a) .You can solve the question in less than 60 seconds.

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Since the answer options involve square roots, we will assume a1 =p=4 and b1 =q=4 (when n=1) Now for n=2,we will get a2= 4*4=16 ; b 2=4*4=16 In our assumption, a2+b 2=32. Check where you are getting 8 among answer options. Answer is option (e) q(pq)(n/2)1 (p+q)= 4*8=32. This won’t take more than more than two minutes.

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26) Let S be the set of all pairs (i,j) where 1≤i

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1) CONSTANT PRODUCT RULE(1/x) & 1/(x+1)

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1) CONSTANT PRODUCT RULE 2) POWER CYCLE 3) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS 4) LAST 2 DIGITS TECHNIQUE 5) SIMLIAR TO DIFFERENT GROUPING ( P&C) 6) APPLICATION OF FACTORIALS 7) GRAPHICAL DIVISION FOR GEOMETRY 8) ASSUMPTION METHOD

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This rule can be applied when we have two parameters whose product is constant, or in other words, when they are inversely proportional to each other. eg)Time x Speed = Distance, Price x consumption = Expenditure, Length x breadth= Area A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other Parameter if the parameters are inversely proportional. Lets see the application with the following examples

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1) A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office? Solution: Assume original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or ½. 1/2 increase in speed will result in 1/3 decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5 hoursà Answer is Distance= 4x1.5=6 km

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2) A 20% increase in price of sugar. Find the % decrease in consumption a family should adopt so that the expenditure remains constant Solution: Here price x consumption= expenditure (constant) Using the technique, 20% (1/5) increase results in 16.66%(1/6) decrease in consumption. Answer=16.66%

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2) POWER CYCLE The last digit of a number of the form ab falls in a particular sequence or order depending on the unit digit of the number (a) and the power the number is raised to (b). The power cycle of a number thus depends on its’ unit digit. Consider the power cycle of 2 2 1=2, 2 5=32 2 2 =4 2 6=64 3 2 =8 2 7=128 4 2 =16 2 8=256 As it can be observed, the unit digit gets repeated after every 4 th power of 2. Hence, we can say that 2 has a power cycle of 2,4,8,6 with frequency 4. This means that, a number of the form 2 4k+1 will have the last digit as 2 2 4k+2 will have the last digit as 4 2 4k+3 will have the last digit as 8 2 4k+4 will have the last digit as 6 (where k=0, 1, 2, 3…) This is applicable not only for 2, but for all numbers ending in 2. Therefore to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0 to 9, which are given below

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Power cycle 0 1 2,4,8,6 3,9,7,1 4,6 5 6 7,9,3,1 8,4,2,6 9,1

Frequency 1 1 4 4 2 1 1 4 4 2

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Unit digit 0 1 2 3 4 5 6 7 8 9

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For example 3) Find the remainder when 375 is divided by 5. 1) Express power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3. 2) Take the power cycle of 3 which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7 Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2. Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in the fastest way possible. For example,

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4) Find the unit digit of 73^4n Put n=1, the problem reduces to 7 3^4, which is 781. Since 81=4k+1, take the first digit in the power cycle of 7, which is 7.

What is the first non zero integer from the right in 83301957 + 8370 1982? a) 3 b) 1 c) 9 d) none of these 8370 1982 will end with more number of zeroes so we need to consider only the first part. Rightmost non-zero integer of the expression will be = unit digit of 8331957 = unit digit of 31957 Since 1957=4k+1, take the first digit in the power cycle of 3, which is 3.

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If N = (13)1! + 2! + 3! + ..+ 13! + (28)1! + 2! + 3!..+ 28! + (32)1! + 2! + 3! + ...+ 32!+ (67) 1! + 2! + 3! + then the unit digit of N is (a) 4 (b) 8 (c) 2 (d) none of these Based on Power Cycle After 4! Every number is of the form 4k+4 , here in all four terms power becomes 4k+1. So taking the first digit from the power cycles of 3,8,2, and 7 we will get the unit digit as i.e 3+8+2+7 = 4. Ans = 0 6)

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3) Useful technique to find the last 2 digits of any expression of the form ab Depending on the last digit of the number in question, we can find the last two digits of that number. We can classify the technique to be applied into 4 categories. TYPE METHOD EXAMPLES 67 1) Numbers ending in 1 The last digit is always 1. 1) 21 =__41(2 * 7=__4) The 2nd last digit = product of tens digit of base * unit digit of the power. In 2167; 2 is the tens digit of base and 7 is the unit digit of power

2) 4187 =__ 81( 4 *7= __8)

2) Number ending with 5

Last two digits: always 25

e.g.) 1555 34 = __ 25

3) Numbers ending in 3, 7, 9

Change the power so that the base ends with 1 and then use the same technique as for those numbers ending with 1.

e.g.) 17288 =(174)72 (taking power 4 as 74 will end in 1. (17 2* 172) 72

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4) 31124 =__ 21( 3 * 4=__2)

=( _89*_89)72 (as last 2 digits of 172=89) = ( _21)72 (as last 2 digits of 89*89=21) Answer =__41( as 2*2=4)

Use the pattern of the number 1024 =2 10 i.e. *2 10 raised to even power ends with 76 and * 210raised to odd power ends with 24.

e.g.) 2 788 = (210) 78 * 28 =

__76 * __56 = __56.

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4) For even numbers (2,4,6,8)

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eg) 34, 7 4 &9 2 all will end in1.

3) 1261 167 =__21 ( 6 * 7= __2)

It is also important to note that,

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1. 76 multiplied by the last 2 digits of any power of 2 will end in the same last 2 digits E.g. 76*04 = 04, 76*08 = 08, 76*16 = 16, 76*32 = 32 2. The last two digits of , x2, (50-x)2 , (50+x)2 , (100-x)2 will always be the same. For example last 2 digits of 122,382,622,882,112 2…. will all be the same. Also, last two digits of 11 2=392=612=89 2 =111 2=139 2=161 2=1892 and so on 3.To find the squares of numbers from 30-70 we can use the following method 7) To find 41 2 Step1 : Difference from 25 will be first 2 digits = 16 Step 2 : Square of the difference from 50 will be last 2 digits = 81 Answer = 1681. 8): To find 432 Step1 : Difference from 25 will be first 2 digits = 18 Step 2 :Square of the difference from 50 will be last 2 digits = 49 Answer = 1849

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4. Combining all these techniques we can find the last 2 digits for any number because every even number can be written as 2* an odd number

4) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS 9) In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B, 80 use brand C, 90 use brand D & 60 use brand E of the same product. What is the minimum possible number of men using all the 5 brands, if all the 100 men use at least one of these brands?

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Sum of the difference from 100 = (100-100) + (100-75)+(100-80)+(100-90)+(100-60) = 95 Again take the difference from 100 = 5 (answer) This method will be explained in detail in the next booklet.

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5) SIMILAR TO DIFFERENT GROUPING IN PERMUTATION & COMBINATION

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All questions in Permutation and Combination fall into 4 categories, and if you master these 4 categories, you can understand all concepts in P&C easily. 1) Similar to Different 2) Different to Similar 3) Similar to Similar 4) Different to Different In this booklet, we will look at the first category; i.e. Similar to Different, which entails the number of ways of dividing ‘n’ identical (similar) things into ‘r’ distinct (different) groups

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a) NO LIMIT QUESTIONS

Let me explain this with an example. Suppose I have 10 identical chocolates to be divided among 3 people. The 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates.

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So let us represent chocolates by blue balls. The straight red lines are used to divide them into parts. So you can see that for dividing into 3 parts, you need only two lines.

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Suppose you want to give 1 st person 1 chocolate, 2nd 3 chocolates and 3rd 6 chocolates. Then you can show it as:

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Suppose you want to give one person 1 chocolate, another person 6 chocolates and another one 3, then it can be represented as:

Now if first person gets 0, second gets 1 and third gets 9 chocolates then it can be represented as:

Now suppose you want to give first person 0, second also 0 and third all of 10 then you can show it like:

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èSo, for dividing 10 identical chocolates among 3 persons you can assume to have 12 (10 balls +2 sticks) things among which ten are identical and rest 2 are same and of one kind. So number of ways in which you can distribute ten chocolates among 3 people is the same in which you can arrange 12 things among which 10 are identical and of one kind while 2 are identical and of one kind which can be done in 12!/(10!2!)

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The above situation is same as finding the number of positive integral solutions of a + b + c = 10. a, b, c is the number of chocolates given to different persons. Or else I can also say how many integral points lie on the line a + b + c = 10 in the 1st quadrant. In both the cases the answer is 12C2.

b) LOWER LIMIT QUESTIONS

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èNow suppose we have a restriction that the groups cannot be empty i.e. in the above example all 3 persons should get at least 1.

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You have to divide ten chocolates among 3 persons so that each gets at least one. So in the start only give them one each. This you will do in just 1 way as all the chocolates are identical. Now, you are left with 7 chocolates and you have to divide them among 3 people in such that way that each gets 0 or more. You can do this easily as explained above using balls and sticks. Number of ways = 9! /(7!2!).

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The above situation is same as finding the number of positive natural number solutions of a + b + c = 10. a, b, c is the number of chocolates given to different persons. Or else I can also say how many integral natural points lie on the line a + b + c = 10 in the 1 st quadrant. In both the cases the answer is 9C2.

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è Now suppose I change the question and say that now you have to divide 10 chocolates among 3 persons in such a way that one gets at least 1, second at least 2 and third at least 3. Its as simple as the last one. First full fill the required condition. Give 1 st person 1, second 2 and third 3 and then divide the left 4 (10–1–2-3) chocolates among those 3 in such a way that each gets at least 1. This is same as arranging 4 balls and 2 sticks which can be done in 6C2 ways.

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The above situation is same as finding the number of positive integral solutions of a + b + c = 10 such that a > = 1, b > = 2, c > = 3. a, b, c is the number of chocolates given to different persons. In this case the answer is 9C2. 10)

Rajesh went to the market to buy 18 fruits in all. If there were mangoes, bananas, apples and oranges for sale then in how many ways could Rajesh buy at least one fruit of each kind? a) 17C3 b) 18C4 c) 21C3 d) 21C4

This is a Grouping type 1 Similar to Distinct question, with a lower limit condition. M+B+A+O=18 Remove one from each group, therefore 4 is subtracted from both sides. The problem changes to M+B+A+O=14 Using our shortcut, The answer is the arrangement of 14 balls and 3 sticks i.e. 17C3

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11)

There are four balls to be put in five boxes where each box can accommodate any number of balls. In how many ways can one do this if: a) Balls are similar and boxes are different 1) 275 2) 70 c) 120 d) 19

When the balls are similar and the boxes are different, it’s a grouping type 1 question A+B+C+D+E=4, where A,B,C,D,E are the different boxes. The number of ways of selection and distribution=8C4=70 The number of non negative integral solutions of x1+x2+x3 ≤ 10 a) 84 b) 286 c) 220 d) none of these

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By non-negative integral solutions, the conditions imply that we can have 0 and natural number values for x1, x2, x3, and x4 To remove the sign ≤ add another dummy variable x4. The problem changes to x1+x2+x3+x4=10 This is an example of grouping type1 (Similar to Distinct) It is the arrangement of 10 balls and 3 sticks. Using the shortcut of balls and sticks, Therefore the answer is 13C3=286

6) APPLICATION OF FACTORIALS

Definition of Factorialà N! = 1*2*3*…*(n-1)*n Eg 1) 5!= 1*2*3*4*5=120 eg 2) 3!=1*2*3=6

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A thorough understanding of Factorials is important because they play a pivotal role not only in understanding concepts in Numbers but also other important topics like Permutation and Combination

I)

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Let us now look at the application of Factorials

Highest power in a factorial or in a product

a)

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Questions based on highest power in a factorial are seen year after year in CAT. Questions based on this can be categorized based on the nature of the number (prime or composite) whose highest power we are finding in the factorial, i.e

Highest power of a prime number in a factorial:

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To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

13) The highest power of 5 in 100!

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100/5=20; 20/5=4; Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24 ALTERNATIVE METHOD 100/5 + 100/52 =20+4=24 (We take upto 52 as it is the highest power of 5 which is less than 100)

b)

Highest power of a composite number in factorial

Factorize the number into primes. Find the highest power of all the prime numbers in that factorial using the previous method. Take the least power.

14) To find the highest power of 10 in 100! Solution: Factorize 10=5*2. 1. Highest power of 5 in 100! =24 2. Highest power of 2 in 100! =97 Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. So take the lesser i.e. 24 is the answer.

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15) Highest power of 12 in 100! Solution: 12=22 *3. Find the highest power of 22 and 3 in 100! First find out the highest power of 2. Listing out the quotients:100/2 = 50; 50/2 = 25; 25/2 = 12; 12/2 = 6; 6/2 = 3; 3/2 = 1 Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97. So highest power of 2 2 = 48 (out of 97 2’s only 48 can make 22) Now for the highest power of 3. 100/3 = 33; 33/3 = 11; 11/3 = 3; 3/3 = 1; Highest power of 3 = 48, Highest power of 12 = 48

II)

Number of zeros in the end of a factorial or a product

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Finding the number of zeroes forms the base concept for a number of application questions. In base 10, number of zeros in the end depends on the number of 10s; i.e. effectively, on the number of 5s In base N, number of zeroes in the end highest power of N in that product

16) Find the number of zeroes in 13! In base 10

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Solution: We need to effectively find the highest power of 10 in 13! = Highest power of 5 in 13! As this power will be lesser. 13/5=2

17) Find the number of zeroes in the end of 15! in base 12.

Number of factors of any factorial

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III)

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Solution: Highest power of 12 in 15! =highest power of 22 *3 in 15! =Highest power of 3 in 15!= 5

There is a technique, which can be used to find the number of factors in a factorial

18) Find the factors of 12!

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STEP 1: Prime factorize 12! i.e. find out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! = 2 10*3 5*52*7*11 STEP2: Use the formula N=am*bn (a, b are the prime factors). Then number of factors= (m+1)(n+1) The number of factors= (10+1)(5+1)(2+1)(1+1)(1+1) =792. Answer=792

Right most non zero integer in a factorial

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19) Find the right most non zero integer in 25! ? OR Find the remainder when 25! is divided by 107?( Both the questions are conceptually the same) There are 6 zeroes at the end of 25!. Effectively we need to find the rightmost non-zero integer in 25!. Number of 2s in 25! = 25/2+12/2+6/2+3/2 = 12+6+3+1=22, Number of 3s in 25! = 25/3 + 8/3 = 8+2=10 Number of 5s in 25! = 25/5+5/1 =6, Number of 7s in 25! =25/7=3 Number of 11s in 25! =2, Number of 13s,17s,19s,23s in 25! =1 Remove 26x5 6. (form the 6 zeroes), Right most non zero digit will be last digit of 2163107311 2.131.171.191.231 = 6.9.3.1.3.7.9.3 = __4. Answer is 4.

A shortcut to solve the above question.

If you know the unit digit in the product 1!*2!*3!*……10! = 8 and Unit digit in the product 11!*12*13*……………..20! = 8.then the answer is in one line as follows. 8*8*1*2*3 = Answer 4.This short cut method takes just 30 seconds.

APPLICATION QUESTION BASED ON FACTORIAL 20) How many natural numbers are there such that their factorials are ending with 5 zeroes? 10! is 1*2*3*4*(5)*6*7*8*9*(2*5). From this we can see that highest power of 5 till 10! is 2. Continuing like this, 10!-14!, highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3). Therefore, from 15! To 19! - The highest power of 5 will be 3.

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20!-24! – Highest Power = 4, In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get highest power of 5 as 6. The answer to the question is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

7) USE OF GRAPHICAL DIVISION IN GEOMETRY

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Lets look at a technique which will help you solve a geometry question in no time 21) ABCD is a square and E and F are the midpoints of AB and BC respectively. Find the ratio of Area( ABCD): Area(DEF)?

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8) ASSUMPTION method

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FIGURE A FIGURE B Lets divide the figure using dotted lines as shown in Figure B. Area of ABCD=100%. Area AEDG=50%. Then Area in shaded region 1(AED)= 25%. Similarly, DCFH=50%. Area in shaded region 2(DCF)=25%. Now Area of EOFB= 25%. .Area of shaded region 3(BEF)=12.5%. Total area outside triangle= 62.5%. Area inside triangle= 10062.5=37.5%. required ratio = 100/37.5 = 8:3

There are at least 10 questions out of 25 in CAT 2007 where you can apply this technique. This involves assuming simple values for the variables in the questions, and substituting in answer options based on those values. Assumption helps to tremendously speedup the process of evaluating the answer as shown below.

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22) k & 2k 2 are the two roots of the equation x 2 – px + q. Find q + 4q 2 + 6pq = a) q2 b) p 3 c) 0 d) 2p3 Solution: Assume an equation with roots 1&2 (k=1) =>p (sum of roots)= 3 and q(product of roots)=2. Substitute in q + 4q 2 + 6pq = 54. Look in the answer options for 54 on substituting values of p=3 and q=2. we get 2p3 = 54.=> Ans = 2p3 .

23) Let ‘x’ be the arithmetic mean and y,z be the two geometric means between any two positive numbers. The value of (y3 + z3) / (xyz) is a) 2 b) 3 c) 1/2 d) 3/2 Assume a GP 1 2 4 8 implies 2 GMs between 1 and 8, i.e. y=2 and z=4. Arithmetic Mean, x =(8+1)/2 = 4.5 .Substitute in (y3 + z3) / (xyz) Answer = (2 3+43)/(2x4x4.5) = 2. NOTE:- Assume a GP 1 1 1 1 then x=1, y=1 z=1. Answer on substitution=2, which will make the calculation even faster, half of the problems in Algebra can be solved using assumption. This is not direct substitution. In the next eg: see how you can use the same technique in an equation question.

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SOME ACTUAL QUESTIONS FROM CAT-2007 SOLVED USING THIS TECHNIQUE

c) q(n/2)(p+q)n/2

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25) Let a 1=p and b1=q, p and q are positive quantities. Define a n=pbn-1, b n=qb n-1 for even n>1 a n=pan-1, b n=qa n-1 for odd n>1. which of the following best describes a n+bn for even n. a) qp (n/2)(p+q) b) q(n/2)(p+q)n (n/2)-1 n/2 d)q(pq) (p+q) e)q(pq)(n/2)-1(p+q)

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24) Consider the set S={2,3,4……2n+1), where n is a positive integer larger than 2007. Define X as the average of odd integers in S and Y as the average of the even integers in S. What is the value of X-Y? a) 1 b) n/2 c)(n+1)/2n d) 2008 e) 0 The question is independent of n, which is shown below. Take n=2. Then S= {2,3,4,5). X= 4 and Y=3. X-Y =1, Take n=3. then S={2,3,4,5,6,7}. X=5 and Y=4. X-Y=1 Hence you can directly mark the answer option (a) .You can solve the question in less than 60 seconds.

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Since the answer options involve square roots, we will assume a1 =p=4 and b1 =q=4 (when n=1) Now for n=2,we will get a2= 4*4=16 ; b 2=4*4=16 In our assumption, a2+b 2=32. Check where you are getting 8 among answer options. Answer is option (e) q(pq)(n/2)1 (p+q)= 4*8=32. This won’t take more than more than two minutes.

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26) Let S be the set of all pairs (i,j) where 1≤i

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